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414 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
+=0.01 moleculesinBaseForm1.0 molecule in Acid Form 1.01 TotalMolecules
==Base Form IonizedFormand Acid Form UnionizedForm
=×
0.01 MoleculesinUnionized Form
1.01 TotalMolecules
0.99%
=×
1.0MoleculeinIonized Form
1.01 TotalMolecules
%
This ratio indicates that for every one molecule that contains the functional group in the
acid (or unionized) form, there are 0.0000079 molecules that contain the functional group
in the base (or ionized) form. The following equations can then be used to correctly calculate the percentage of the molecules that are ionized and the percentage that are unionized.
rcentinIonized Form
100%
=
rcentinUnionized Form
=
100% 99.001
The question asks for the percent that are ionized, so the correct answer is
Review Questions
1. Answers provided in the grid below.
Drug (pKa Value) Name of Functional Group Acidic/Basic
Carvedilol (7.8) Secondary amine Basic
Ketoprofen (5.94) Carboxylic acid Acidic
Dasolampanel (3.93) Tetrazole Acidic
Dasolampanel (6.73) Secondary amine Basic
Haloperidol (8.6) Tertiary amine Basic
2. The acidic, basic, or neutral environments are provided below.
Saliva (pH = 6.4) Stomach (pH = 2) Duodenum (pH = 5.4) Plasma (pH = 7.4) Urine (pH = 5.7)
Acidic Acidic Acidic Basic (nearly
neutral)
Acidic
.
3. Predominant ionization states are provided below.
Drug (pKa value) Saliva (pH = 6.4) Stomach (pH = 2) Duodenum (pH = 5.4) Plasma (pH = 7.4)
Carvedilol (7.8) Ionized Ionized Ionized Ionized
Ketoprofen (5.94) Ionized Unionized Unionized Ionized
Dasolampanel (3.93) Ionized Unionized Ionized Ionized
Dasolampanel (6.73) Ionized Ionized Ionized Unionized
Haloperidol (8.6) Ionized Ionized Ionized Ionized
4. Tolbutamide contains an acidic functional group (sulfonylurea; pKa 5.4), and it is predomi-
nantly ionized when the environmental pH > pKa. As stated in the chapter text, the Rule of
Nines can be used in lieu of the Henderson-Hasselbalch equation if the difference between
the pH and pKa is an integer, as it is in this scenario. Using this rule, we can calculate the
approximate percent ionization in each of these environments (duodenum pH = 5.4; saliva
pH = 6.4; plasma pH = 7.4) by simply determining the magnitude of the difference between
the pH and pKa. In the duodenum, the pH = pKa; therefore, 50% of the carboxylic acid is
ionized at any given time and 50% is unionized. In the saliva, the difference between the pH
APPENDIX - ANSWERS TO CHAPTER QUESTIONS 415
[UnprotonatedForm]
[Basic Form]
aa
[UnprotonatedForm]
[UnprotonatedForm]
[ProtonatedForm]
+=2,931molecules in IonizedForm1.0 molecule in UnionizedForm2,932 TotalMolecules
=×
2,931Molecules in IonizedForm
2,932Total Molecules
=×
1MoleculeinUnionized Form
2,932Total Molecules
%
[UnprotonatedForm]
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and the pKa values is 1; therefore, 90% of the carboxylic acid functional groups is ionized
and 10% is unionized. In the plasma, the difference between the pH and the pKa values is 2;
therefore, 99% of the carboxylic acid functional groups is ionized and 1% is unionized.
5. Part A: Qualitative approach (predominantly ionized or unionized?)
Dasolampanel contains a tetrazole (acidic; pKa = 3.93) in an environment pH = 7.4. For
acids, if pH > pKa, then the functional group is predominantly ionized.
Dasolampanel contains a secondary amine (basic; pKa = 6.73) in an environment pH = 7.4.
For basic functional groups, if the pH > pKa, then the functional group is predominantly
unionized.
Part B: Quantitative approach (% ionized and % unionized?)
Answer for the Acidic Functional Group (Dasolampanel):
The form of the Henderson-Hasselbalch equation used for acidic functional groups is
shown below.
=+ =+
HpKlog
[ProtonatedForm]
or pH pK log
[AcidicForm]
When using this equation to determine whether the functional group is predominantly ion-
ized or unionized, we need to know the pKa of the functional group (tetrazole in dasolampanel pKa = 3.93) and the pH of the environment (plasma pH = 7.4).
=+
43.93log
=
log
[ProtonatedForm]
The fact that the number on the left is positive indicates that the tetrazole is predominantly
in its unprotonated (basic, ionized) form at pH = 7.4.
To determine the percentage of ionized and unionized drug in the plasma, we need to cal-
culate the antilog of both sides of this equation. In this case, the antilog of 3.47 is 2,951.
This means that for every one molecule that is protonated (unionized tetrazole), 2,951 molecules are unprotonated (ionized tetrazole).
Calculation of the percentage of ionized drug and unionized drug at pH = 7.4 requires one
additional step.
rcentofMolecules in IonizedForm
=
100% 99.96%
rcentofMolecules in UnionizedForm
=
100% 0.04
Answer for the Basic Functional Group (Dasolampanel):
The form of the Henderson-Hasselbalch equation used for basic functional groups is show
below.
=+
HpKlog
a
[ProtonatedForm]
416 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
[UnprotonatedForm]
[ProtonatedForm]
[UnprotonatedForm]
+=1moleculeinIonized Form 4.86 moleculesinUnionized Form 5.86 TotalMolecules
=×
4.86 MoleculesinUnionized Form
5.86 TotalMolecules
%
=×
1MoleculeinIonized Form
5.86 TotalMolecules
%
When using this equation to determine whether the functional group is predominantly ion-
ized or unionized, we need to know the pKa of the functional group (secondary amine in
dasolampanel pKa = 6.73) and the pH of the environment (plasma pH = 7.4).
=+
46.73log
=
67 log
[ProtonatedForm]
The fact that the number on the left is positive indicates that the secondary amine is pre-
dominantly in its unprotonated (basic, unionized) form at pH = 7.4.
To determine the percentage of ionized and unionized drug in the plasma, we need to calcu-
late the antilog of both sides of the equation. In this case, the antilog of 0.67 is 4.68. This
means that for every one molecule that is protonated (ionized secondary amine), there are
4.86 molecules that are unprotonated (unionized secondary amine).
Calculation of the percentage of ionized drug and unionized drug at pH = 7.4 requires one
additional step.
rcentofMolecules in UnionizedForm
100% 82.9
rcentofMolecules in IonizedForm
=
100% 17.1
6. Based on ionization only (which allows for ion–dipole interactions with water) ketoprofen
is the only one of the four drugs that is unionized in the duodenum. Although this molecule
can participate in other types of interactions with water (e.g., H-bonding) it cannot participate in ion–dipole interactions because it is not ionized in that environment.
Based on ionization only (which allows for ion–dipole interactions with water) dasolam-
panel is the only one of the four drugs that has two ionized functional groups in the duodenum. This molecule is able to interact with water via both the ionized secondary amine and
the ionized tetrazole.
7. To determine if a drug can participate in an ionic interaction with the ionized arginine residue found within the active site of COX-1, we first need to determine what the ionized form
of arginine looks like at pH = 7.4. Using the Henderson-Hasselbalch equation to determine
if this functional group is predominantly ionized in the plasma (pH = 7.4), we learn that
99.9991% of the molecules of arginine are ionized in that environment (see structure of
ionized arginine side chain below).
=
For a drug to participate in an ionic interaction with the side chain of this arginine residue,
it must contain a functional group that is negatively charged in the same environment.
APPENDIX - ANSWERS TO CHAPTER QUESTIONS 417
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Because we have already determined that ketoprofen is predominantly ionized at pH = 7.4
(via the carboxylic acid), it stands to reason that ketoprofen can participate in this critical
ionic interaction with the ionized arginine residue.
Haloperidol contains a basic tertiary amine (piperidine) and, when ionized, is positively
charged. As we determined in Question 3, haloperidol is predominantly ionized in this environment (pH = 7.4). Because an ionic interaction requires that an ion pair form between
one negatively and one positively charged functional group, it stands to reason that the
positively charged tertiary amine (piperidine) in haloperidol cannot participate in an ionic
interaction with the positively charged side chain of the arginine residue.
8. Acidic and basic functional groups are identified below.
Functional
Group Name
Secondary
amine
Phenol Acidic
Acidic or
Basic
Basic
(pK
= 9-11)
a
= 9-10)
(pK
a
Saliva
(pH = 6.4)
Ionized Ionized Ionized Ionized Ionized
Unionized Unionized Unionized Unionized Unionized
Stomach
(pH = 2)
Duodenum
(pH = 5.4)
Plasma
(pH = 7.4)
Urine
(pH = 5.7)
There are no physiologic environments in which batefenterol is predominantly in its union-
ized form. For the drug to be absorbed via passive diffusion, we can see that in the stomach
the phenol is primarily in its unionized form; however, only a very small fraction of the
secondary amine is in its unionized form in that environment at any given time. We know
that an equilibrium exists between the unionized and ionized form of the drug and that the
equilibrium is re-established as the unionized drug is absorbed, so we might anticipate that
little by little batefenterol will be absorbed from the stomach.
9. Part A: Acidic and basic functional groups and their respective pKa values are shown below.
Part B: The hydrochloride salt form of the drug is ionized (secondary amine is in its proto-
nated form) and, therefore, capable of participating in an ion–dipole interaction with water.
This enhances the solubility of the drug in an aqueous solution.
418 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
[UnprotonatedForm]
[Basic form]
aa
[UnprotonatedForm]
[UnprotonatedForm]
+=1.1Molecules in IonizedForm1.0 Molecule in UnionizedForm2.1 TotalMolecules
=×
1.1Molecules in IonizedForm
2.1Total Molecules
52.38%
=×
1MoleculeinUnionized Form
2.1Total Molecules
[UnprotonatedForm]
[UnprotonatedForm]
[UnprotonatedForm]
Part C:
Answer for Acidic Functional Group (Carboxylic Acid)
The form of the Henderson-Hasselbalch equation used for acidic functional groups is shown
below.
=+ =+
HpKlog
[ProtonatedForm]
or pH pK log
[Acidic]
When using this equation to determine whether the functional group is predominantly ion-
ized or unionized, we need to know the pKa of the functional group (carboxylic acid in moxifloxacin pKa = 6.3) and the pH of the environment (formulation pH = 7.4).
=+
46.3 log
=
1log
[ProtonatedForm]
[ProtonatedForm]
The fact that the number on the left is positive indicates that the carboxylic acid is pre-
dominantly in its unprotonated (basic, ionized) form at pH = 7.4.
To determine the percentage of ionized and unionized drug in the formulation, we need to
calculate the antilog of both sides of this equation. In this case, the antilog of 1.1 is 12.59.
This means that for every one molecule that is protonated (unionized carboxylic acid;
acidic), there are 12.59 molecules that are unprotonated (ionized carboxylic acid; basic).
Calculation of the percentage of ionized drug and unionized drug at pH = 7.4 requires one
additional step.
rcentofMolecules in IonizedForm
100%
=
rcentofMolecules in UnionizedForm
100% 47.62%
Answer for Basic Functional Group (Secondary Amine/Piperazine)
The form of the Henderson-Hasselbalch equation used for basic functional groups is show
below.
=+
HpHlog
a
[ProtonatedForm]
When using this equation to determine whether the functional group is predominantly ion-
ized or unionized, we need to know the pKa of the functional group (secondary amine in
moxifloxacin pK
= 9.3) and the pH of the environment (formulation pH = 7.4).
a
=+
49.3 log
=1.9log
[ProtonatedForm]
[ProtonatedForm]
The fact that the number on the left is negative indicates that the secondary amine is pre-
dominantly in its protonated (acidic, ionized) form at pH = 7.4.
=
APPENDIX - ANSWERS TO CHAPTER QUESTIONS 419
+=1MoleculeinIonized Form 0.012589 MoleculesinUnionized Form 1.012589 TotalMolecules
=×
1MoleculeinIonized Form
1.012589 TotalMolecules
%
=×
0.012589 MoleculesinUnionized Form
%
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To determine the percentage of ionized and unionized drug in the formulation, we need
to calculate the antilog of both sides of the equation. In this case, the antilog of −1.9 is
0.012589. This means that for every one molecule that is unprotonated (unionized secondary amine), there are 0.012589 molecules that are protonated (ionized secondary amine).
Calculation of the percentage of ionized drug and unionized drug at pH = 7.4 requires one
additional step.
rcentofMolecules in IonizedForm
rcentofMolecules in UnionizedForm
1.012589 TotalMolecules
10. Part A: Functional groups are matched with the appropriate pKa values below.
Part B: Ionization states and explanations are provided below.
=
100% 99.87
100% 0.13
=
Drug (pKa Value) Stomach (pH = 1.8) Urine (pH = 6.1) Cell (pH = 7.4)
Cefotaxime (3.4) Primarily unionized Primarily ionized Primarily ionized
Nitrofurantoin (7.1) Primarily unionized Primarily unionized Primarily ionized
Atenolol (9.6) Primarily ionized Primarily ionized Primarily ionized
Ezetimibe (10.2) Primarily unionized Primarily unionized Primarily unionized
Cefotaxime contains an acidic carboxylic acid. At a urine pH of 6.1 and a cellular pH of 7.4,
the environment (i.e., the pH) is more basic than the functional group (i.e., the pH > pK
acidic functional group is primarily ionized in a basic environment. In contrast, at a stomach
pH of 1.8, the environment is more acidic than the functional group (i.e., the pH < pKa). An
acidic functional group is primarily unionized in an acidic environment.
Nitrofurantion contains an acidic imide group. At a stomach pH of 1.8 and a urinary pH of 6.1,
the environment (i.e., the pH) is more acidic than the functional group (i.e., the pH < pKa).
An acidic functional group is primarily unionized in an acidic environment. At a cellular pH
). An
a
420 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
[BaseForm]
[AcidForm]
[BaseForm]
==39.8
[BaseForm]
39.8
[BaseForm]
+=39.8 MoleculesinBaseForm1.0 Molecule in Acid Form 40.8 TotalMolecules
==Base Form IonizedForm, andAcidFormUnionized Form forthisfunctionalgroup
=×Pe
39.8 MoleculesinIonized Form
40.8 TotalMolecules
00%
=PercentinIonized Form 97.5%
of 7.4, the environment is slightly more basic than the functional group (i.e., the pH > pKa).
An acidic functional group is primarily ionized in a basic environment.
Atenolol contains a basic secondary amine with a pKa value that is greater than any of the
above three environments. In all of these situations, the environment is more acidic than
the functional group (i.e., the pH < pKa). This basic functional group is primarily ionized in
all three of these acidic environments.
Ezetimibe contains an acidic phenol with a pKa value that is greater than any of the above
three environments. In all of these situations, the environment is more acidic than the functional group (i.e., the pH < pKa). This acidic functional group is primarily unionized in all
three of these acidic environments.
11. Part A: Functional groups are matched with the appropriate pKa below.
Part B: Calculations for percent ionization at a pH = 6.2 are provided below.
For the carboxylic acid, the pKa = 4.6 and the pH = 6.2. Using the Henderson-Hasselbalch
equation gives the following:
=+
24.6 log
Solving the equation provides a ratio of [Base Form]/[Acid Form].
=
6log
[AcidForm]
[AcidForm]
or
1
[AcidForm]
Thus, for every one molecule that contains this functional group in the acid form, there are
39.8 molecules that contain this functional group in the base form. Because the functional
group is acidic, the base form is equal to the ionized form and the acid form is equal to the
unionized form.
rcentinIonized Form
1
APPENDIX - ANSWERS TO CHAPTER QUESTIONS 421
[BaseForm]
[AcidForm]
[BaseForm]
==0.006
[BaseForm]
0.0063
[BaseForm]
+=0.0063Molecules in Base Form 1.0MoleculeinAcidForm1.0063Total Molecules
==Base Form UnionizedFormand Acid Form IonizedFormfor this functional group
=×Pe
1MoleculeinIonized Form
1.0063Total Molecules
00%
=PercentinIonized Form 99.4%
[BaseForm]
[AcidForm]
[UnprotonatedForm]
[ProtonatedForm]
[30]
0]
−=p
[30]
)7.83
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For the primary amine, the pKa = 8.4 and the pH = 6.2. Using the Henderson-Hasselbalch
equation gives the following.
=+
28.4 log
Solving the equation provides a ratio of [Base Form]/[Acid Form].
3
[AcidForm]
=2.2log
[AcidForm]
or
1
[AcidForm]
Thus, for every one molecule that contains this functional group in the acid form, there are
0.0063 molecules that contain this functional group in the base form. Because the functional group is basic, the base form is equal to the unionized form, and the acid form is equal
to the ionized form.
rcentinIonized Form
1
12. The functional group with the pKa value of 8.2 is a tertiary amine as shown below. The nitro
group adjacent to the other nitrogen atoms is electron withdrawing and greatly decreases
the basicity and pKa value for this functional group.
We can use the Henderson-Hasselbalch equation to solve this problem. Because the ion-
ized form of a basic functional group can also be designated as its protonated form or its
conjugate acid form, either of the following equations can be used.
=+
HpKlog
=+
rpHpKlog
a
a
Therefore, if 70% of this functional group is ionized:
=+
=+ =+
H8.2 log
H8.2 log
[70]
8.2(0.37
[7
Because a pH of 7.83 does not exist physiologically, this percent ionization could only occur
in an exogenously prepared solution.
422 BASIC CONCEPTS IN MEDICINAL CHEMISTRY
CHAPTER 5
Structural Analysis Checkpoint
Checkpoint Drug 1: Venetoclax
1. The potassium salt of venetoclax is shown below.
Part A: In Chapter 3, we identified five ionizable functional groups. Four of these five func-
tional groups are basic while one is acidic. To form a salt, it is necessary to combine an acid
and a base in a solution. In this case, an inorganic potassium salt was formed, so it can
be concluded that the potassium ion came from KOH, a strong inorganic molecule. The
addition of KOH to a solution containing venetoclax causes a significant increase in the pH
of the solution. All four basic functional groups are primarily unionized, while the acidic
sulfonamide is primarily ionized. Under these conditions, the negatively charged sulfonamide reacts with the positively charged potassium to form the salt shown in the answer to
Question 1.
Part B: Shown below is a diolamine salt of venetoclax. Another base that could be used to
make a salt with the sulfonamide is tromethamine.
Part C: Water-soluble organic salts are able to provide enhanced solvation and dissolution
as compared with inorganic salts. The inorganic potassium ion is able to form ion-dipole
bonds with water (as the ion). This is also true with the nitrogen atom of diolamine salt;
however, the two hydroxyl groups allow additional hydrogen bonds with water, thus further enhancing water solubility beyond what is possible with the potassium salt.
APPENDIX - ANSWERS TO CHAPTER QUESTIONS 423
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2. The structure of venetoclax contains a large number of functional groups that can form
hydrogen bonds or ion–dipole interactions with water. These have been highlighted below.
Please note that the tertiary amine and the sulfonamide (boxed) is primarily ionized in the
pH of the small intestine and would participate in ion–dipole interactions with water. These
functional groups provide sufficient water solubility for venetoclax to dissolve in the gastrointestinal (GI) tract. The remaining functional groups (aromatic and alicyclic rings, alkyl
chains, and halogens) provide sufficient lipid solubility to allow venetoclax to traverse the
GI membrane once it is dissolved.
3. As compared with the original structure of venetoclax, the addition of a hydroxyl group
enhances water solubility because this functional group can form hydrogen bonds with
water. Additionally, the addition of a hydroxyl group allows for the formation of water- and
lipid-soluble ester prodrugs. As discussed in the chapter, esterases are ubiquitous throughout the human body, allowing the release of the active drug in the liver, the blood stream,
the GI tract, or the target tissue.
4.
The chlorine group is lipid soluble; therefore, removing it and replacing it with a hydrogen
atom would decrease the overall lipid solubility, or increase the overall water solubility,
of venetoclax. Additionally, because the chlorine group is electron withdrawing through
induction, the aromatic ring would no longer be electron deficient. This could affect charge
transfer interactions and other interactions between the aromatic ring and its biological target. A full explanation of these types of interactions is provided in the next chapter. Additionally, electron withdrawing groups on aromatic rings decrease the metabolic
oxidation of the ring. This is discussed in Chapter 8. Finally, because a hydrogen atom is
smaller than a chlorine atom, the para position of this aromatic ring would be less sterically
hindered.
Checkpoint Drug 2: Elamipretide
1. Part A: Elamipretide can be formulated as a sodium or potassium salt (phenol) or as a
hydrochloride or hydrobromide salt (guanidine and primary amine).
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